Question

Find the largest 4 digit number which when divided by 42 and 24 leaves 33 and 15 as remainders respectively.

Answer 1

Observe 42 - 33 = 9 & 24 - 15 = 9

Required number will be 9 more than a multiple of LCM(42,24).

Now LCM(42,24)= 168

Required number will be N = 168k + 9

N = 1008 + 9 ( for k = 6)

= 1017 ( smallest 4 digit number )

N = 9912 + 9 ( for k = 59 )

= 9921 ( largest 4 digit number )

hope it helps!!